What does this expression mean, and why does it compile? [duplicate]

Question

After a typo, the following expression (simplified) compiled and executed:

if((1 == 2) || 0 (-4 > 2))
  printf("Hello");

of course, the 0 shouldn't be there.

Why does it compile, and what does the expression mean?

The original (simplified) should look like this:

if((1 == 2) || (-4 > 2))
  printf("Hello");

none of this does compile:

if((1 == 2) || true (-4 > 2))
  printf("Hello");

if((1 == 2) || 1 (-4 > 2))
  printf("Hello");

if((1 == 2) || null (-4 > 2))
  printf("Hello");

Answer 1

It looks like this is a Visual C++ extension to support a particular 'no function defined' idiom. From the warning C4353 page:

// C4353.cpp
// compile with: /W1
void MyPrintf(void){};
#define X 0
#if X
   #define DBPRINT MyPrint
#else
   #define DBPRINT 0   // C4353 expected
#endif
int main(){
    DBPRINT();
}

the intention being that DBPRINT is a no-op. The warning suggests #define DBPRINT __noop instead, using VC's __noop extension instead.

If you view the assembly listing for your output you'll see the second clause is omitted, even in debug mode.

Answer 2

Guess it was interpreted as

if((1 == 2) || NULL (-4 > 2))
  printf("Hello");

where NULL is a function-pointer, by default returning int... What at actually happens in runtime is platform-dependent